imc

Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,b{r}^2,b{r}^3.$ We are given that $$\begin{array}{rl}a+b& =57,\\ a+d+br& =60,\\ a+2d+b{r}^2& =91,\end{array}$$ and we wish to find $a+3d+b{r}^3.$ Subtracting the first equation from the second and the second equation from the third, we get $$\begin{array}{rl}d+b(r-1)& =3,\\ d+br(r-1)& =31.\end{array}$$ Subtract these results, we get $$b(r-1{)}^2=28.$$ Note that either $b=28$ or $b=7.$ We proceed with casework: If $b=28,$ then $r=2,a=29,$ and $d=-25.$ The arithmetic sequence is $29,4,-21,-46,$ arriving at a contradiction. If $b=7,$ then $r=3,a=50,$ and $d=-11.$ The arithmetic sequence is $50,39,28,17,$ and the geometric sequence is $7,21,63,189.$ This case is valid. Therefore, The answer is $a+3d+b{r}^3=17+189=\boxed{206}.$